| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Linear relationship μ = kσ |
| Difficulty | Standard +0.8 This question requires standardization of normal distributions and working with inverse normal tables in part (a), then in part (b) solving a system involving both the constraint 2μ = 3σ² and a probability condition P(Y > 3μ/2) = 0.0668, requiring algebraic manipulation of the relationship between parameters. The combination of probability work with simultaneous equations involving mean and variance makes this more challenging than routine normal distribution questions. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X > \mu + 2k) = 0.2\) or \(P(X < \mu - 2k) = 0.2\) | M1 | Any correct tail probability statement; also implied by \(\pm 0.84\) seen |
| \(\frac{\mu + 2k - \mu}{6} = 0.8416\) or \(\frac{\mu - 2k - \mu}{6} = -0.8416\) | M1 A1 | Standardising using \(\mu\) and 6, setting equal to \(z\) where \(0.8 < \ |
| \(k = 2.5248...\) awrt \(2.52\) | A1 | awrt 2.52 (allow 2.525) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P\!\left(Y > \frac{3}{2}\mu\right) \Rightarrow P\!\left(Z > \frac{\frac{3}{2}\mu - \mu}{\sigma}\right) \Rightarrow P\!\left(Z > \frac{\frac{1}{2}\mu}{\sigma}\right)\) | M1 | Standardising using \(\frac{3}{2}\mu\), \(\mu\) and \(\sigma\) |
| Substituting \(\mu = \frac{3}{2}\sigma^2\) giving \(P\!\left(Z > \frac{3}{4}\sigma\right)\) or setting up two equations in \(\mu\) and \(\sigma\) | M1 | Substitution of \(\mu = \frac{3}{2}\sigma^2\) into standardisation or setting up two equations |
| \(\frac{3}{4}\sigma = 1.5\) or \(\frac{1}{2}\sqrt{\frac{3\mu}{2}} = 1.5\) or \(3\sigma^2 = 6\sigma\) | M1 | Expression for \(\sigma\) only or \(\mu\) only used with \(\pm 1.5\) |
| \(\mu = 6\) only, \(\sigma = 2\) only | A1 A1 | Final A1: \(\mu=6\) and \(\sigma=2\) must reject any other values if found |
## Question 8:
### Part (a):
| $P(X > \mu + 2k) = 0.2$ or $P(X < \mu - 2k) = 0.2$ | M1 | Any correct tail probability statement; also implied by $\pm 0.84$ seen |
| $\frac{\mu + 2k - \mu}{6} = 0.8416$ or $\frac{\mu - 2k - \mu}{6} = -0.8416$ | M1 A1 | Standardising using $\mu$ and 6, setting equal to $z$ where $0.8 < \|z\| < 0.9$; A1 for fully correct standardisation with $\|z\| = 0.8416$ or better |
| $k = 2.5248...$ awrt $2.52$ | A1 | awrt 2.52 (allow 2.525) |
### Part (b):
| $P\!\left(Y > \frac{3}{2}\mu\right) \Rightarrow P\!\left(Z > \frac{\frac{3}{2}\mu - \mu}{\sigma}\right) \Rightarrow P\!\left(Z > \frac{\frac{1}{2}\mu}{\sigma}\right)$ | M1 | Standardising using $\frac{3}{2}\mu$, $\mu$ and $\sigma$ |
| Substituting $\mu = \frac{3}{2}\sigma^2$ giving $P\!\left(Z > \frac{3}{4}\sigma\right)$ or setting up two equations in $\mu$ and $\sigma$ | M1 | Substitution of $\mu = \frac{3}{2}\sigma^2$ into standardisation or setting up two equations |
| $\frac{3}{4}\sigma = 1.5$ or $\frac{1}{2}\sqrt{\frac{3\mu}{2}} = 1.5$ or $3\sigma^2 = 6\sigma$ | M1 | Expression for $\sigma$ only or $\mu$ only used with $\pm 1.5$ |
| $\mu = 6$ only, $\sigma = 2$ only | A1 A1 | Final A1: $\mu=6$ and $\sigma=2$ must reject any other values if found |\begin{enumerate}
\item The random variable $X$ is normally distributed with mean $\mu$ and variance 36
\end{enumerate}
Given that
$$\mathrm { P } ( \mu - 2 k < X < \mu + 2 k ) = 0.6$$
(a) find the value of $k$
The random variable $Y$ is normally distributed with mean $\mu$ and standard deviation $\sigma$ Given that
$$2 \mu = 3 \sigma ^ { 2 } \quad \text { and } \quad \mathrm { P } \left( \mathrm { Y } > \frac { 3 } { 2 } \mu \right) = 0.0668$$
(b) find the value of $\mu$ and the value of $\sigma$
\hfill \mbox{\textit{Edexcel S1 2024 Q8 [9]}}